My Math Forum Constructing a formula

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 November 15th, 2007, 02:12 AM #1 Newbie   Joined: Nov 2007 Posts: 3 Thanks: 0 Constructing a formula I always knew how to do this, until yesterday, my mind just went blank. I'm writing a computer program which will ask questions on constructing a formula... I need to know how to write down the working to find this out. This has been bothering me. Thanks to all who contributes, Garry.
 November 15th, 2007, 02:51 AM #2 Senior Member   Joined: May 2007 Posts: 402 Thanks: 0 What kind of formula and what kind of questions?
 November 15th, 2007, 03:03 AM #3 Newbie   Joined: Nov 2007 Posts: 3 Thanks: 0 Oh yeah, oops. Forgot to say what kind of formula. How do you find out that formula. M ---- 1 2 3 4 5 6 7 8 Y ----- 6 9 12 15 18 Y = 2M + 5
 November 15th, 2007, 03:29 AM #4 Senior Member   Joined: May 2007 Posts: 402 Thanks: 0 Ok, so what you need is a line. First, let me say that you can make a line go through exactly 2 random points, or any number of specific points. So, the information you gave is too much - you only need two points. So, let's say you have these points: T1(x1,y1) T2(x2,y2) and you need an equation of the form: y(x)=a*x+b So, plugging in the two points you get: y(x1)=y1=a*x1+b y(x2)=y2=a*x2+b ------------------------ subtract the second one from the first y1-y2=a*x1+b-a*x2-b y1-y2=a(x1-x2) ------------------------ solve for 'a' a=(y1-y2)/(x1-x2) ------------------------ plug a into the first (or the second) formula y1=(y1-y2)/(x1-x2)*x1+b ------------------------ solve for b b=y1-(y1-y2)/(x1-x2)*x1 b=y1-(y1*x1-y2*x1)/(x1-x2) b=(y1*x1-y1*x2-y1*x1+y2*x1)/(x1-x2) b=(y2*x1-y1*x2)/(x1-x2) ------------------------ plug that and 'a' into y(x)=a*x+b y(x)=(y1-y2)/(x1-x2)x+(y2*x1-y1*x2)/(x1-x2) y(x)=(y1-y2)/(x1-x2)x+(y2*x1+y2*x2-y2*x2-y1*x2)/(x1-x2) y(x)=(y1-y2)/(x1-x2)x+(y2*x1-y2*x2)/(x1-x2)+(y2*x2-y1*x2)/(x1-x2) y(x)=(y1-y2)/(x1-x2)x+y2+(y2-y1)/(x1-x2)x2 y(x)-y2=(y1-y2)/(x1-x2)(x-x2) ------------------------ or if we had plugged it into the second equation y(x)-y1=(y1-y2)/(x1-x2)(x-x1) but both of them yield the same result!
 November 15th, 2007, 03:43 AM #5 Newbie   Joined: Nov 2007 Posts: 3 Thanks: 0 Made no sense to me. Not what my teacher taught me. It was way simpler. I'm sorry I didn't understand. That must of took a while, but thank you for your help anyway.
November 15th, 2007, 05:41 AM   #6
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Joined: Dec 2006

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I'm not sure I would have done it quite like that. Since we are trying to find a linear line, all we need are two points, like you said, but I would find the equation a little differently: (Maybe "milin" could explain further why he used the method he did, as he must have had a reason).

Quote:
 M ---- 1 2 3 4 5 6 7 8 Y ----- 6 9 12 15 18
y = AM + b (standard linear equation form)

First, we find the slope between the first two points, (1,6) and (2,9):

Slope = (9-6)/(2-1) = 3

y = 3M + b (new equation)

Now, plug in any point (We'll use 1,6) and find the value of 'b':

6 = 3(1) + b

3 = b

y = 3M + 3 Final equation of the line which goes through all the points you originally mentioned.

(I'm not sure where you got the "y = 2M + 5" equation from, but it does not work)

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